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(C) The given reaction is a nucleophilic substitution reaction of a tertiary alkyl halide,$(C_6H_5)_3C-Cl$,which proceeds via an $SN_1$ mechanism.In a

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(C) The given reaction is a nucleophilic substitution reaction of a tertiary alkyl halide,$(C_6H_5)_3C-Cl$,which proceeds via an $SN_1$ mechanism.In an $SN_1$ reaction,the rate-determining step is the formation of the carbocation,which depends only on the concentration of the alkyl halide.The rate law for this reaction is: $	ext{Rate} = k[(C_6H_5)_3C-Cl]$.Since the rate is directly proportional to the concentration of the alkyl halide,the graph of rate versus $[(C_6H_5)_3C-Cl]$ should be a straight line passing through the origin,as shown in option $C$.

The graph which represents the following reaction is :
$(C_6H_5)_3C-Cl \xrightarrow{OH^{-} / \text{Pyridine}} (C_6H_5)_3C-OH$

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Major product which you expect in the above reaction is

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The graph which represents the following reaction is :$(C_6H_5)_3C-Cl \xrightarrow{OH^{-} / \text{Pyridine}} (C_6H_5)_3C-OH$